# FP2: Taylors series

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#1
any ideas how I'd tackle this?
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#2
do I use this formula and make a = 0?
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1 year ago
#3
(Original post by Maths&physics)
do I use this formula and make a = 0?
Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.
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1 year ago
#4
(Original post by Maths&physics)
any ideas how I'd tackle this?
You seek a truncated expansion of , where it can be expanded in powers of as:  implies and , this means that To get the fourth term, you can differentiate the ODE and see what is. The second term should be fairly obvious.
Last edited by RDKGames; 1 year ago
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#5
(Original post by mqb2766)
Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.
what is ODE?
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1 year ago
#6
Sorry, the differential equation. Just differentiate it with respect to x again.
ODE is Ordinary Differential Equation - you don't need to know the term, but can look it up if you want.

(Original post by Maths&physics)
what is ODE?
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1 year ago
#7
(Original post by Maths&physics)
what is ODE?
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!
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#8
(Original post by RDKGames)
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!
is that a term in the specification because I've never heard it before. 0
1 year ago
#9
(Original post by Maths&physics)
is that a term in the specification because I've never heard it before. No.
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1 year ago
#10
(Original post by Maths&physics)
is that a term in the specification because I've never heard it before. I don't know... check your spec online and CTRL+F for "ODE" to see if it appears or not...

But TBH, I would be very surprised (and disappointed) if the resources you use for this theory do not abbreviate the topic to ODE because it's an extremely standard acronym.
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#11
(Original post by mqb2766)
No.
(Original post by RDKGames)
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation

I.e. the equation with the derivative that they give you!
I got for the third term... (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
Last edited by Maths&physics; 1 year ago
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1 year ago
#12
(Original post by Maths&physics)
I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
Last edited by mqb2766; 1 year ago
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1 year ago
#13
(Original post by Maths&physics)
I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
That equation looks dodgy... how did you arrive at that??
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#14
(Original post by mqb2766)
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
(Original post by RDKGames)
That equation looks dodgy... how did you arrive at that??
I meant: (d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next? I don't understand how to get dx/dt?

I sub x = 0, and I get: (d^3)y/d(t^3) = -(dx/dt)
Last edited by Maths&physics; 1 year ago
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1 year ago
#15
(Original post by Maths&physics)
I meant: (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next?
That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?
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#16
(Original post by RDKGames)
That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?
(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)
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1 year ago
#17
(Original post by Maths&physics)
(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)
Nope... [second line is fine tho...]
You should really lay it out properly.  When , we get: You know every single value here apart from ... so rearrange for it.
Last edited by RDKGames; 1 year ago
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#18 that's what I meant! (d^3)x/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0
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#19
(Original post by Maths&physics) that's what I meant! (d^3)x/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0
(Original post by mqb2766)
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
I'm so sorry, I confused my notations with the tutorial I watching.
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